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# If zy < xy < 0, is | x - z | + |x| = |z|?

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If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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17 Sep 2010, 12:10
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If zy < xy < 0, is | x - z | + |x| = |z|?

(1) z < x
(2) y > 0
[Reveal] Spoiler: OA
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If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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17 Sep 2010, 15:58
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This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If $$zy<xy<0$$ is $$|x-z|+|x| = |z|$$

Look at the inequality $$zy<xy<0$$:

We can have two cases:

A. If $$y<0$$ --> when reducing we should flip signs and we'll get: $$z>x>0$$.
In this case: as $$z>x$$ --> $$|x-z|=-x+z$$; as $$x>0$$ and $$z>0$$ --> $$|x|=x$$ and $$|z|=z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$-x+z+x=z$$ --> $$0=0$$, which is true.

And:

B. If $$y>0$$ --> when reducing we'll get: $$z<x<0$$.
In this case: as $$z<x$$ --> $$|x-z|=x-z$$; as $$x<0$$ and $$z<0$$ --> $$|x|=-x$$ and $$|z|=-z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$x-z-x=-z$$ --> $$0=0$$, which is true.

So knowing that $$zy<xy<0$$ is true, we can conclude that $$|x-z|+|x| = |z|$$ will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that $$z<x$$, hence we have case B.

Statement (2) says that $$y>0$$, again we have case B.

Hope it helps.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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20 Jun 2012, 05:05
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rgtiwari wrote:
If zy < xy < 0, is | x - z | + |x| = |z|?

(1) z < x
(2) y > 0

You can solve such questions easily by re-stating '< 0' as 'negative' and '> 0' as 'positive'.

zy < xy < 0 implies both 'zy' and 'xy' are negative and zy is more negative i.e. has greater absolute value as compared to xy. Since y will be equal in both, z will have a greater absolute value as compared to x.

When will zy and xy both be negative? In 2 cases:
Case 1: When y is positive and z and x are both negative.
Case 2: When y is negative and z and x are both positive.

Question:
Is | x - z | + |x| = |z| ?
Is | x - z | = |z| - |x| ?
Is | z - x | = |z| - |x| ? (Since | x - z | = | z - x |)
We re-write the question only for better understanding.

Now think, when will | z - x | = |z| - |x| ? It happens when x and z have the same sign. In case both have the same sign, they get subtracted on both sides so you get the same answer. In case they have opposite signs, they get added on LHS and subtracted on RHS and hence the equality doesn't hold.

So if we can figure whether both x and z have the same sign, we can answer the question.

As we saw above, in both case 1 and case 2, x and z must have the same sign. This implies that the equality must hold and you don't actually need the statements to answer the question. You can answer it without the statements (this shouldn't happen in actual GMAT).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 04 Jun 2010 Posts: 112 Concentration: General Management, Technology Schools: Chicago (Booth) - Class of 2013 GMAT 1: 670 Q47 V35 GMAT 2: 730 Q49 V41 Followers: 15 Kudos [?]: 269 [4] , given: 43 Re: Absolute values [#permalink] ### Show Tags 17 Sep 2010, 12:33 4 This post received KUDOS Hi, rgtiwari Draw a number line and put 0 in the middle of it. now look at the data you are given. zy<xy<0. we don't know if y is positive or negative. if y>0 than z is on the left of x which is on the left of zero => z<x and both z & x are originally negative. the opposite if y<0, x will be on the right of zero and z will be on the right of x. => z>x. and both are originally positive. in both cases it seems that the equation is correct. but don't bother to check that just look at the extra data and continue from there. If you find it hard to understand with variables just use numbers instead. now let's look at (1) it tells you that z<x so we know that this is the first case only => the equality is definitely correct (just use numbers that maintain the data in the first case) - Sufficient. (2) it tells you that y>0. again we are at the first case. - Sufficient. pay attention that this is not a coincidence that the extra data given leads you in (1) and (2) to the same conclusion. if it doesn't you should suspect whether it's D or not. Hope that helps. _________________ Consider Kudos if my post helped you. Thanks! -------------------------------------------------------------------- My TOEFL Debrief: http://gmatclub.com/forum/my-toefl-experience-99884.html My GMAT Debrief: http://gmatclub.com/forum/670-730-10-luck-20-skill-15-concentrated-power-of-will-104473.html Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 101 Kudos [?]: 948 [2] , given: 43 Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink] ### Show Tags 18 Sep 2012, 00:44 2 This post received KUDOS rgtiwari wrote: If zy < xy < 0, is | x - z | + |x| = |z|? (1) z < x (2) y > 0 Since zy < 0 and xy < 0, both z and x have opposite sign to y, so they must be either both positive or both negative. In other words, we know that xz > 0. (1) Given that z < x, when both z and x are negative, |z - x| + |x| = -z + x + (-x) = -z = |z| TRUE z and x cannot be both positive, because then y would be negative, and from zy < xy we would obtain that z > x. Sufficient. (2) Knowing that y > 0, we can deduce that both z and x are negative. In addition, from zy < xy it follows that z < x, and we are in the same case as above. Sufficient. Answer D _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7380 Location: Pune, India Followers: 2292 Kudos [?]: 15166 [2] , given: 224 Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink] ### Show Tags 20 Dec 2012, 20:50 2 This post received KUDOS Expert's post eaakbari wrote: If | z - x |= |z| - |x| assuming x & z have same sign | x - z |= |z| - |x| ; Since | x - z | = | z - x | Implies |z| - |x| = |x| - |z| which is defn not true z = 1 , x = 2 Plugging, we get - 1 = 1 ?!? Have I misunderstood something? Before discussing this part, I have discussed in my post that absolute value of z must be greater than absolute value of x. If absolute value of z is not greater than absolute value of x, then | z - x |= |z| - |x| does not hold when x and z have the same sign. Since |z| must be greater than |x|, | x - z | = |x| - |z| does not hold. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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18 Sep 2010, 00:47
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Got the approach. Thanks Rafi and Bunuel.
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Re: If zy<xy<0 is |x-z| + |x| = |z|? 1. z<x>0 2. [#permalink]

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17 Sep 2012, 15:16
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successstory wrote:
If zy<xy<0 is |x-z| + |x| = |z|?

1. z<x>0
2. y>0

why not a?

The question surprisingly does not require either statements for it to be true.

|x-z| + |x| = |z| can be rearranged as |x-z| = |z|-|x|. Now since we have an equal sign( opposed to an inequality, in which we would have to check both sides to make sure they are both positive or both negative to square (and flip)), we square both sides to get (|x-z|)^2 = (|z|-|x|)^2 which yields ===> xz=|x||z|

So our question becomes: Is xz=|x||z|?

Well this can only be true if (x>0 and z>0) OR (x<0 and z<0). In other words x and z must both be the same sign for the above statement to be true.

Now we are given zy<xy<0 as a fact. Two cases arise (+)(-)<(+)(-)<0 or (-)(+)<(-)(+)<0. Notice in both cases x and z are always the same sign!!!!! This statement is true from the gecko. Hence whatever unnecessary statement GMAC tells us will be sufficient. This is an usual problem you wont see again.
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26 Nov 2012, 18:54
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Is this really a 600 level question? Given its time consuming nature, seems more like a 750 level one.

Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happen then the answer should be D.

If $$zy<xy<0$$ is $$|x-z|+|x| = |z|$$

Look at the inequality $$zy<xy<0$$:

We can have two cases:

A. If $$y<0$$ --> when reducing we should flip signs and we'll get: $$z>x>0$$.
In this case: as $$z>x$$ --> $$|x-z|=-x+z$$; as $$x>0$$ and $$z>0$$ --> $$|x|=x$$ and $$|z|=z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$-x+z+x=z$$ --> $$0=0$$, which is true.

And:

B. If $$y>0$$ --> when reducing we'll get: $$z<x<0$$.
In this case: as $$z<x$$ --> $$|x-z|=x-z$$; as $$x<0$$ and $$z<0$$ --> $$|x|=-x$$ and $$|z|=-z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$x-z-x=-z$$ --> $$0=0$$, which is true.

So knowing that $$zy<xy<0$$ is true, we can conclude that $$|x-z|+|x| = |z|$$ will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that $$z<x$$, hence we have case B.

Statement (2) says that $$y>0$$, again we have case B.

Hope it helps.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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20 Dec 2012, 09:32
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VeritasPrepKarishma wrote:
Now think, when will | z - x | = |z| - |x| ? It happens when x and z have the same sign.

If | z - x |= |z| - |x| assuming x & z have same sign
| x - z |= |z| - |x| ; Since | x - z | = | z - x |

Implies |z| - |x| = |x| - |z|
which is defn not true
z = 1 , x = 2
Plugging, we get
- 1 = 1 ?!?

Have I misunderstood something?
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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08 Feb 2013, 12:18
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The question is not good.
The target question "If zy < xy < 0, is | x - z | + |x| = |z|?"
tells us zy<xy<0, that is, z and x have the same sign,
Thus, target question could be rephrased as "| x - z | = |z|-|x| ?"=> "|z|>|x| ?"
The condition zy<xy<0 could be rephrased as |zy|>|xy|>0 => |z|>|x| , which is already enough to solve the question.
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15 May 2013, 06:30
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from given

y(z-x) < 0 thus z is not = x and x is not equal to zero

if we square the 2 sides of the question

we get

x^2 + z^2 - 2xz +2x^2 - 2xz +x^2 = z^2 this boiles down to is 4x ( x-z) = 0 ? the answer is yes if x = 0 or x=z and no if we can know for sure that neother x=z nor x = 0

and this is given in the question stem.... neither givens are needed ..........as the answer is for sure NO
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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26 Feb 2017, 03:22
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OreoShake wrote:
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If $$zy<xy<0$$ is $$|x-z|+|x| = |z|$$

Look at the inequality $$zy<xy<0$$:

We can have two cases:

A. If $$y<0$$ --> when reducing we should flip signs and we'll get: $$z>x>0$$.
In this case: as $$z>x$$ --> $$|x-z|=-x+z$$; as $$x>0$$ and $$z>0$$ --> $$|x|=x$$ and $$|z|=z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$-x+z+x=z$$ --> $$0=0$$, which is true.

And:

B. If $$y>0$$ --> when reducing we'll get: $$z<x<0$$.
In this case: as $$z<x$$ --> $$|x-z|=x-z$$; as $$x<0$$ and $$z<0$$ --> $$|x|=-x$$ and $$|z|=-z$$.

Hence in this case $$|x-z|+|x|=|z|$$ will expand as follows: $$x-z-x=-z$$ --> $$0=0$$, which is true.

So knowing that $$zy<xy<0$$ is true, we can conclude that $$|x-z|+|x| = |z|$$ will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that $$z<x$$, hence we have case B.

Statement (2) says that $$y>0$$, again we have case B.

Hope it helps.

why in case A, have you expanded |x-z| as -x+z, but |x| and |y| as x and y?? shouldnt |x| and |y| be expanded as -x and -y as well?

A. If $$y<0$$ --> when reducing we should flip signs and we'll get: $$z>x>0$$.

Since $$z>x$$ --> x - z < 0, thus $$|x-z|=-(x-z) =-x+z$$;
Since $$x>0$$ and $$z>0$$ --> $$|x|=x$$ and $$|z|=z$$.
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If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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29 Nov 2015, 21:19
|x-z| + |x| = |z| ?

|x-z| = (absolute value of the sum of x and z, if x and z have different signs) OR (absolute value of the difference between x and z, if x and z have the same sign)

|x-z| = |z| - |x|? Reformulated questions:

Q1) Do x and z have the same sign?(|z| - |x| represents a difference not a sum, so in |x-z|, x and z should have the same sign for this to be true) and
Q2) is |z| > |x|(The difference |z| - |x| is positive, therefore |z| should be > |x|)?

Information given in the question:

zy < xy < 0: gives two possibilities:

Possibility 1) y is negative and x and z are positive and the z > x
Possibility 2) y is positive and x and z are negative and z < x, |z| > |x|

Statement (1) z < x, therefore possibility 2 is true, both x and z are negative, and the equality is correct.
Statement (2) y > 0, therefore possibility 2 is true, both x and z are negative, and the equality is correct.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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03 Dec 2015, 05:54
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If zy < xy < 0, is | x - z | + |x| = |z|?

(1) z < x
(2) y > 0

We get y(z-x)<0, is |x-z|=|z|-|x|?, zx>0 and z>x? if we modify the question.
There are 3 variables (x,y,z) and 1 equation in the original condition,
2 more equations in the given conditions, so there is high chance (C) will be the answer.
But condition 1=condition 2 answering the question 'no' and sufficient.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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03 Dec 2015, 05:59
rgtiwari wrote:
If zy < xy < 0, is | x - z | + |x| = |z|?

(1) z < x
(2) y > 0

The question although doe slook menacing but can be solved easily.

You are given zy<xy ---> y (z-x)<0

Per statement 1, z<x --> x-z>0 ---> |x-z| = x-z , thus the LHS of the question becomes = 2x-z and this is definitely not = z .

Per statement 2, y > 0. This also leads to the same information as in statement 1 as from y (z-x)<0 --> if y > 0 then the only case possible is for z<x and as shown in statement 1 , this is sufficient to answer the question.

D is the thus the correct answer.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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17 Mar 2016, 01:21
data in question stem : zy<xy<0

look at options A) z<x that means 'y' is positive and option B) says the same thing y>0
that means both statements eventually say the same thing
so the answer must be either E or D
take y>0 case and see , we can establish a CONFORM YES condition answer must be D
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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07 Apr 2016, 08:15
Experts please help. What level would you rate this problem? I am currently doing a DS set and got this question right in around 2 minutes but then read that its a sub-600 level some where. Please confirm?
Re: If zy < xy < 0, is | x - z | + |x| = |z|?   [#permalink] 07 Apr 2016, 08:15

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